Simple Question About Residues/Poles/Zeros/Singularities I'm having a little bit of trouble with residues.
If we have the $f(z)=\left(\frac{\cos(z)-1}{z}\right)^2$ at $z_0=0$, we have a zero of order 2 in the numerator and a zero of order 2 in the denominator, which means we have the residue = 0 correct?
Would we only consider the denominator to have a pole if the denominator contained a $(z-z_0)$? If so, isn't it already in that form?
edit:
For my problem I think I understand now. 
$$f(z)=\left(\sum_{n=1}^\infty\frac{z^{2n-1}(-1)^n}{(2n)!}\right)^2$$ Will have the lowest power of $z$ being $z^2$ so all $b_n$ are zero and thus its a removable singularity and this its residue is zero.
 A: In an expression like $1/(z-z_0)$, it is not the denominator that has a pole; it is the quotient.
To say that $f(z)$ has a pole at $z=z_0$ means that $f(z)\to\infty$ as $z\to z_0$.  Anything about a pole occurring at places where the denominator is $0$ and the numerator is not is a consequence of the definition rather than a part of the definition.  If the numerator is holomorphic and not $0$ at $z_0$ and the denominator is holomorphic and $0$ at $z_0$, then the quotient has a pole at $z_0$.  But if the numerator and denominator are both holomorphic and both equal to $0$ at $z_0$, then there is more work to do to decide whether the quotient approaches $\infty$ at that point.
In the expression $f(z)=\left(\dfrac{\cos(z)-1}{z}\right)^2$, there is a zero of order $2$ in the denominator and a zero of order $4$, not $2$, in the numerator.  The numerator is
$$
\left( -\frac{z^2}2 + \frac{z^4}{24} - \frac{z^6}{720} + \cdots \right)^2 = \frac{z^4} 4 + \text{higher-degree terms}.
$$
A: You are correct. If we define $f(0)=0$, then this is an entire function (analytic on all of $\mathbb{C}$) which has no poles or singularities for finite $z$.
